Advanced Mathematical Methods for Scientists and Engineers Episode 3 Part 4 pot

Consider a second order, linear, homogeneous differential equation:Show that P00− Q0+ R = 0 is a necessary and sufficient condition for this equation to be exact.. Thus the general solut

The matrix of eigenvectors and its inverse is

and that the function of a matrix in Jordan canonical form is

We want to compute eJt so we consider the function f (λ) = eλt, which has the derivative f0(λ) = t eλt Thus

The exponential matrix is

e−Aτg(τ ) dτ

x = eAt

Z t 0

We differentiate and multiply by t to obtain a second order coupled equation for x1 We use (15.4) to eliminatethe dependence on x2.

t2x001 + tx01 = atx01+ btx02

t2x001 + (1 − a)tx01 = b(cx1+ dx2)

t2x001+ (1 − a)tx01− bcx1 = d(tx01− ax1)

t2x001 + (1 − a − d)tx01+ (ad − bc)x1 = 0Thus we see that x1 satisfies a second order, Euler equation By symmetry we see that x2 satisfies,

3 Suppose that λ = α is an eigenvalue of multiplicity 2 If λ = α has two linearly independent eigenvectors, a and

b then atα and btα are linearly independent solutions If λ = α has only one linearly independent eigenvector,

a, then atα is a solution We look for a second solution of the form

x = xitαlog t + ηtα.Substituting this into the differential equation yields

αxitα−1log t + xitα−1+ αηtα−1 = Axitα−1log t + Aηtα−1

We equate coefficients of tα−1log t and tα−1 to determine xi and η

(A − αI)xi = 0, (A − αI)η = xi

These equations have solutions because λ = α has generalized eigenvectors of first and second order.

Note that the change of independent variable τ = log t, y(τ ) = x(t), will transform (15.4) into a constantcoefficient system

dy

dτ = AyThus all the methods for solving constant coefficient systems carry over directly to solving (15.4) In the case ofeigenvalues with multiplicity greater than one, we will have solutions of the form,

xitα, xitαlog t + ηtα, xitα(log t)2 + ηtαlog t + ζtα, ,analogous to the form of the solutions for a constant coefficient system,

1 0

1 1

x

The characteristic polynomial of the matrix is

χ(λ) =

1 − λ 0

1 1 − λ

.There is only one linearly independent eigenvector, which we choose to be

a =01



One solution of the differential equation is

x1 =0

1

t

We look for a second solution of the form

.The solution is determined only up to an additive multiple of a We choose

η =10

.Thus a second linearly independent solution is

The general solution of the differential equation is

x = c1

01



t + c2

01



t log t +1

0

t



Method 2 Note that the matrix is lower triangular

x0 1

x02



= 1t

By substituting the solution for x1 into (15.5), we obtain an uncoupled equation for x2.

tx2 = c1log t + c2

x2 = c1t log t + c2tThus the solution of the system is

x = c1

t

t log t

+ c20t

,

which is equivalent to the solution we obtained previously

Consider a second order, linear, homogeneous differential equation:

Show that P00− Q0+ R = 0 is a necessary and sufficient condition for this equation to be exact

Hint, Solution

Exercise 16.2

Determine an equation for the integrating factor µ(x) for Equation16.1

Exercise 16.4

On what intervals do the following problems have unique solutions?

1 xy00+ 3y = x

2 x(x − 1)y00+ 3xy0+ 4y = 2

Particular Solutions Any function, yp, that satisfies the inhomogeneous equation, L[yp] = f (x), is called aparticular solution or particular integral of the equation Note that for linear differential equations the particular solution

is not unique If yp is a particular solution then yp+yh is also a particular solution where yh is any homogeneous solution.The general solution to the problem L[y] = f (x) is the sum of a particular solution and a linear combination of thehomogeneous solutions

y = yp+ c1y1+ · · · + cnyn.Example 16.2.1 Consider the differential equation

y00− y0 = 1

You can verify that two homogeneous solutions are ex and 1 A particular solution is −x Thus the general solution is

y = −x + c1ex+c2.Exercise 16.5

Suppose you are able to find three linearly independent particular solutions u1(x), u2(x) and u3(x) of the second orderlinear differential equation L[y] = f (x) What is the general solution?

Hint, Solution

Real-Valued Solutions If the coefficient function and the inhomogeneity in Equation 16.2 are real-valued, thenthe general solution can be written in terms of real-valued functions Let y be any, homogeneous solution, (perhapscomplex-valued) By taking the complex conjugate of the equation L[y] = 0 we show that ¯y is a homogeneous solution

as well

L[y] = 0L[y] = 0

y(n)+ pn−1y(n−1)+ · · · + p0y = 0

¯(n)+ pn−1¯(n−1)+ · · · + p0y = 0¯

L [¯y] = 0

For the same reason, if yp is a particular solution, then yp is a particular solution as well.

Since the real and imaginary parts of a function y are linear combinations of y and ¯y,

y = yp+ c1y1 + · · · + cnynwhere yp is a particular solution, L[yp] = f , and the yk are linearly independent homogeneous solutions, L[yk] = 0 If the coefficient functions and inhomogeneity are real-valued, then the general solution can be written in terms of real-valued functions.

16.3 Transformation to a First Order System

Any linear differential equation can be put in the form of a system of first order differential equations Consider

y(n)+ pn−1y(n−1)+ · · · + p0y = f (x)

We introduce the functions,

y1 = y, y2 = y0, , , yn = y(n−1).The differential equation is equivalent to the system

y01 = y2

y02 = y3

= .

yn0 = f (x) − pn−1yn− · · · − p0y1.The first order system is more useful when numerically solving the differential equation

Example 16.3.1 Consider the differential equation

Result 16.4.1 Let aij(x), the elements of the matrix A, be differentiable functions of x Then

x x22x 4x3

= x4− 2x3+ 4x4− 2x3

= 5x4− 4x3

16.4.2 The Wronskian of a Set of Functions.

A set of functions {y1, y2, , yn} is linearly dependent on an interval if there are constants c1, , cn not all zero suchthat

c1y1+ c2y2+ · · · + cnyn = 0 (16.3)identically on the interval The set is linearly independent if all of the constants must be zero to satisfy c1y1+· · · cnyn = 0

on the interval

Consider a set of functions {y1, y2, , yn} that are linearly dependent on a given interval and n − 1 times entiable There are a set of constants, not all zero, that satisfy equation16.3

differ-Differentiating equation 16.3 n − 1 times gives the equations,

y1 y2 yn

y10 y20 y0n

. .

y1(n−1) y2(n−1) yn(n−1)

... of y and ¯y,

y = yp+ c1y1 + · · · + cnynwhere yp is a particular... x22x 4x3< /small>

= x4< /sup>− 2x3< /small>+ 4x4< /sup>− 2x3< /small>

= 5x4< /sup>− 4x3< /small>

16 .4. 2 The Wronskian... solution, L[yp] = f , and the yk are linearly independent homogeneous solutions, L[yk] = If the coefficient functions and inhomogeneity